Question

Codeine contains a basic nitrogen atom that can be protonated to give the conjugate acid of codeine. Calculate the pH of `1.1x10^-3`M of codeine if the pKa of the conjugate acid is 8.21?

Answer 1

We interrogate the reaction...

`C_18H_21NO_3(aq) + H_2O(l) rarrC_18H_22NO_3^(+) + HO^-`

Explanation:

And since `pK_b+pK_a=pK_w`...

`pK_b=14-pK_a=14-8.21;` then `K_b=10^(-5.79)=1.621xx10^-6`

And so we got our familiar equilibrium reaction....

`K_b=1.621xx10^-6=([C_18H_22NO_3^(+)][HO^-])/([C_18H_21NO_3(aq)])`

And if we set `[C_18H_22NO_3^(+)]=x`...then

`1.621xx10^-6=(x^2)/(1.1xx10^-3-x)`

And IF we assume that `1.1xx10^-3">>"x`..then..

`x_1=sqrt(1.621xx10^-6xx1.1xx10^-3)=4.22xx10^-5`..and we plug that first approx. back into the equation to see how `x` evolves...

`x_2=sqrt(1.621xx10^-6xx(1.1xx10^-3-4.22xx10^-5))=4.14xx10^-5`....and again..

`x_3=sqrt(1.621xx10^-6xx(1.1xx10^-3-4.14xx10^-5))=4.14xx10^-5`..

And since the values have converged I am prepared to accept this as the true value....i.e.

`x=[HO^-]=[C_18H_22NO_3^+]=4.14xx10^-5*mol*L^-1`

But we want `pH` of this solution....well, we can find `pOH=-log_10(4.14xx10^-5)=4.38`...and `pH=14-pOH=9.62`.

The solution is slightly basic as we would anticipate for a weak base in water.

Confused yet? But all I have done is to make successive approximation, ahd, hopefully, avoided arithmetric error....